Is $$\frac{d}{dx} \left(\mathbf{A}_{3\times3}\mathbf{U}(x)_{3\times3}\mathbf{b}_{3\times1}\right) = \mathbf{A}_{3\times3}\frac{d\mathbf{U}(x)_{3\times3}}{dx}\mathbf{b}_{3\times1}$$
Where $\mathbf{A}$ and $\mathbf{b}$ are independent of $x$.
This Wikipedia link states that it is true, but can anyone prove/confirm this?
You can write the entries $(\mathbf A \mathbf U)_{i,j}$ of $\mathbf A \mathbf U$ as $$(\mathbf A\mathbf U)_{i,j} = \sum_k\mathbf A_{i,k}\mathbf U_{k,j}$$ Similarly, the entries of $\mathbf A \mathbf U \mathbf b$ are $$(\mathbf A \mathbf U \mathbf b)_i =\sum_j \sum_k \mathbf A_{i,k}\mathbf U_{k,j} \mathbf b_j$$ So the entries of $\frac{d}{dx}\mathbf A \mathbf U \mathbf b$ are: $$\frac{d}{dx}(\mathbf A \mathbf U \mathbf b)_i = \sum_j \sum_k \mathbf A_{i,k}\frac{d \mathbf U_{k,j}}{dx} \mathbf b_j = \left(\mathbf A \frac{d\mathbf U}{dx} \mathbf b\right)_i$$ So $\frac{d}{dx}\mathbf A \mathbf U \mathbf b = \mathbf A \frac{d\mathbf U}{dx} \mathbf b$.